Another Trignometric Series

First Of All We Assume An Equation With Negative Index

${ \left( 1+x \right) }^{ -n }$

Where $\quad n \quad \epsilon \quad {R}^{+}$

${ \left( 1+x \right) }^{ -n }$ = $1 - nx + \dfrac{n(n+1}{2} x^{2} - \dfrac{n(n+1)(n+2)}{1.2.3} \cdots$

And Comparing With the Given Equation,

$1 - \dfrac{1}{2} cosx + \dfrac{1.3}{2.4} cos2x - \dfrac{1.3.5}{2.4.6} cos3x \cdots$

We Get,

$nx = \dfrac{1}{2}cosx$

And

$\dfrac{n(n+1)}{2}x^{2} = \dfrac{1.3}{2.4}cos2x$

Solving Them We Get

$n = \dfrac{1}{2}$

And

Substituting In The Third Term To $\textbf{ Verify The Consistency.}$

Then The Required Solution Is

$(1+e^{i\theta})^{\dfrac{-1}{2}}$ = $1 - \dfrac{1}{2} e^{i\theta} + \dfrac{1.3}{2.4}(e^{i\theta})^{2}- \dfrac{1.3.5}{2.4.6}(e^{i\theta})^{3} \cdots$

Using De Moivre's formula That

$(\cos x + i \sin x)^\alpha= \cos (\alpha x) + i \sin (\alpha x)$

We Get,

$(1+e^{i\theta})^{\dfrac{-1}{2}}$ = $1 - \dfrac{1}{2}(\cos (x) + i \sin (x)) + \dfrac{1.3}{2.4}(\cos (2x) + i \sin (2x))- \dfrac{1.3.5}{2.4.6}(\cos (3x) + i \sin (3x)) \cdots$

$Re{(1+e^{i\theta})^{\dfrac{-1}{2}}}$ = $1 - \dfrac{1}{2} cosx + \dfrac{1.3}{2.4} cos2x - \dfrac{1.3.5}{2.4.6} cos3x \cdots$

$\textbf{Where Re is the Real Part Of the Complex Number}$

So

$f(x) = Re{(1+e^{i\theta})^{\dfrac{-1}{2}}}$

$f(\dfrac{\pi}{3}) = Re{(1+e^{i\dfrac{\pi}{3}})^{\dfrac{-1}{2}}}$

Upon Simplifying The Above Equation We get,

$f(\dfrac{\pi}{3}) =\sqrt{\dfrac{2cos^2(\dfrac{\pi}{12})}{cos(\dfrac{\pi}{6})}}$

$f(\dfrac{\pi}{3}) = 3^{\dfrac{-1}{4}}.\cos(\dfrac{\pi}{12})$ = $\boxed{0.733944\cdots}$

We have:

$\frac{1}{\sqrt{1+x}}=1-\frac{1}{2}x+\frac{1*3}{2*4}x^2-\frac{1*3*5}{2*4*6}x^3...$

The required expression is then the real part of what one gets upon making the substitution

$x= e^{ix}$

Let $z=\frac{1}{\sqrt{1+e^{ix}}}$ .

Then,

$\Re(z) =\frac{z+\bar{z}}{2}$

Where $\bar{z}=\frac{1}{\sqrt{1+e^{-ix}}}$

This upon further simplification gives:

$\Re(z) =\sqrt{\frac{2cos^2(\frac{x}{4})}{cos(\frac{x}{2})}}$

Now just plug in $x=\frac{\pi}{3}$ .

Consider the binomial expansion of $\dfrac 1{\sqrt{1+x}}$ :

$\begin{aligned} (1+x)^{-\frac 12} & = 1 - \frac 12 x + \frac {\frac 12 \times \frac 32}{2!} x^2 - \frac {\frac 12 \times \frac 32 \times \frac 52}{3!} + \cdots \\ & = 1 - \frac 12 x + \frac {1\times 3}{2\times 4}x^2 - \frac {1\times 3 \times 5}{2\times 4 \times 6} x^3 + \cdots \\ \frac 1{\sqrt{1+e^{xi}}} & = 1 - \frac 12 e^{xi} + \frac {1\times 3}{2\times 4}e^{2xi} - \frac {1\times 3 \times 5}{2\times 4 \times 6} e^{3xi} + \cdots & \small \color{#3D99F6} \text{By Euler's formula: }e^{\theta i} = \cos \theta + i \sin \theta \\ \Re \left(\frac 1{\sqrt{1+e^{xi}}}\right) & = 1 - \frac 12 \cos x + \frac {1\times 3}{2\times 4}\cos 2x - \frac {1\times 3 \times 5}{2\times 4 \times 6} \cos 3x + \cdots & \small \color{#3D99F6} \text{where }\Re (\cdot) \text{ denotes the real part of a complex number.} \end{aligned}$

$\begin{aligned} \implies f (x) & = \Re \left(\frac 1{\sqrt{1+e^{xi}}}\right) \\ f\left(\frac \pi 3\right) & = \Re \left(\frac 1{\sqrt{1+e^{\frac \pi 3i}}}\right) = \Re \left(\frac 1{\sqrt{\frac 32 + \frac {\sqrt 3}2i}}\right) = \Re \left(\frac 1{\sqrt{\sqrt 3\left(\frac {\sqrt 3}2 + \frac 12i\right)}}\right) \\ & = \Re \left(\frac {e^{-\frac \pi {12}i}}{\sqrt[4]3}\right) = \Re \left(\frac {\cos \frac \pi{12}-i \sin \frac \pi{12}}{\sqrt[4]3}\right) = = \frac {\cos \frac \pi{12}}{\sqrt[4]3} \approx 0.7339 \end{aligned}$

Therefore, $\left \lfloor 10000 f\left(\frac \pi 3\right) \right \rfloor = \boxed{7339}$ .